This question is about equilibrium - AQA - A-Level Chemistry - Question 5 - 2021 - Paper 2

The structure of the diester C2H4O2 can be represented as:

  O
  ||
HO-C-(CH2)-C-OH
  ||
  O

This structure shows the diester with the hydroxyl groups that participate in the reaction.

Using the equilibrium concentrations from Table 3:

• Amount of diester: $0.971$ mol
• Amount of acid: $0.452$ mol
• Amount of diol: $0.273$ mol

The equilibrium expression is given by:

$K_c = \frac{[\text{Acid}]^2 [\text{Diol}]}{[\text{Diester}][\text{Water}]^2}$ Substituting the known values:

$0.161 = \frac{(0.452)^2(0.273)}{(0.971)([\text{Water}]^2)}$ Solving for [Water], we calculate:

1. Calculate the numerator: ${(0.452^2)(0.273)}$
2. Rearranging the equation: $[\text{Water}]^2 = \frac{0.452^2 * 0.273}{0.161 * 0.971}$
3. Extracting the square root gives the final amount of water in mol.